There are infinitely many positive integers $k$ which satisfy the equation
\[\cos^2 (k^2 + 6^2)^\circ = 1.\]Enter the two smallest solutions, separated by commas.
Explanation: Note that $\cos^2 \theta = 1$ if and only if $\theta$ is a multiple of $180^\circ.$  Thus, we seek $k$ so that
\[k^2 + 36 = 180n\]for some nonnegative integer $n.$  Then
\[k^2 = 180n - 36 = 36 (5n - 1).\]Hence, $k$ must be a multiple of 6.  We see that $k = 6$ does not work, but $k = \boxed{12}$ and $k = \boxed{18}$ work, so these are the two smallest solutions.